Evaluating definite integral

1. Mar 24, 2013

Saitama

1. The problem statement, all variables and given/known data
$$\int_{0}^{2\pi} \frac{dx}{1+e^{\sin x}}$$

2. Relevant equations

3. The attempt at a solution
Let $I=\int_{0}^{2\pi} \frac{dx}{1+e^{\sin x}}$

Since $\int_{a}^{b}f(x) dx=\int_{a}^{b} f(a+b-x)dx$
Hence,
$$I=\int_{0}^{2\pi} \frac{dx}{1+e^{-\sin x}}=\int_{0}^{2\pi} \frac{e^{\sin x}dx}{1+e^{\sin x}}$$

$$2I=\int_{0}^{2 \pi}dx$$
$$I=\pi$$

This is the right answer but when I try it with the following method it turns out to be zero.

Let $\sin x=t$ or $\cos xdx=dt$. Multiply the numerator and denominator by $\cos x$ & $\cos x=\sqrt{1-t^2}$
The integral transforms to
$$I=\int_{0}^{0} \frac{dt}{\sqrt{1-t^2} (1+e^t)}$$
Both the upper and lower limits are zero. The value of I should be zero. What have I done wrong?

Any help is appreciated. Thanks!

2. Mar 24, 2013

ShreyasR

Yo Naruto fan :)
Last year i was confused with the same problem... when a trigonometric function is applied to 0 or 2∏, we get the same result... As trigonometric functions are periodic. You can draw a graph and u can see that the area under the curve is not 0. This implies that you can not substitute a trigonometric function like "sin x" with a variable like "t" when the difference between the upper and lower limits is greater than ∏/2... It doesn't appear wrong... but it is something like to prove LHS=RHS by multiplying both sides of an equation with zero...!
Hope this helped!

3. Mar 24, 2013

jfgobin

Hello there,

Be careful with:

$\cos x = \sqrt{1- \sin^{2} x}$​

In particular, for $0 \le x \le 2\,\pi$:

$\cos x = \left \{ \begin{array}{ll} \sqrt{1- \sin^{2} x} & 0 \le x \le \pi/2\\ -\sqrt{1- \sin^{2} x} & \pi/2 < x \le 3\,\pi/2 \\ \sqrt{1- \sin^{2} x} & 3\,\pi/2 < x \le 2\,\pi \end{array}\right .$​

In your case, split the integral in 3 cases, and go ahead.

4. Mar 24, 2013

Curious3141

JFGobin has provided the correct explanation, but you should note that once you split the integral up, you don't really save any work - you basically need to do the same thing you did the first time to get the result (plus at one point you need to evaluate $\displaystyle \int_{-1}^{1}\frac{1}{\sqrt{1-t^2}}dt$, which is easy, but still extra work). You do get the same answer though (I checked).

5. Mar 24, 2013

Saitama

Thank you Curious and jfgobin!

The given integral can be separated into three integrals.
$$I=\int_{0}^{\pi/2} \frac{\cos x}{\sqrt{1-\sin^2x}(1+e^{\sin x})}dx-\int_{\pi/2}^{3\pi/2} \frac{\cos x}{\sqrt{1-\sin^2x}(1+e^{\sin x})}dx+\int_{3\pi/2}^{2\pi} \frac{\cos x}{\sqrt{1-\sin^2x}(1+e^{\sin x})}dx$$
Using the substitution $\sin x=t$.
$$I=\int_{0}^{1} \frac{dt}{\sqrt{1-t^2}(1+e^t)}+\int_{-1}^{1} \frac{dt}{\sqrt{1-t^2}(1+e^t)}+\int_{-1}^{0} \frac{dt}{\sqrt{1-t^2}(1+e^t)}$$
$$I=I_1+I_2+I_3$$

I did a change of variable in $I_3$, I substituted $t=-x$ or $dt=-dx$. The integral $I_3$ transforms to
$$I_3=\int_{1}^{0} \frac{-dx}{\sqrt{1-x^2}(1+e^{-x})}$$
which is equivalent to
$$I_3=\int_{0}^{1} \frac{e^tdt}{\sqrt{1-t^2}(1+e^{t})}$$

Now I can add $I_3$ and $I_1$. I end up with this:
$$I=\int_{0}^{1} \frac{dt}{\sqrt{1-t^2}}+\int_{-1}^{1} \frac{dt}{\sqrt{1-t^2}(1+e^t)}$$
But my expression doesn't include the integral mentioned by Curious3141.

6. Mar 24, 2013

Dick

$I_1+I_3=I_2$. So if $I_1+I_3=\int_{0}^{1} \frac{dt}{\sqrt{1-t^2}}$ then $I=2\int_{0}^{1} \frac{dt}{\sqrt{1-t^2}}=\int_{-1}^{1} \frac{dt}{\sqrt{1-t^2}}$ (because the integral is symmetric in t).

7. Mar 24, 2013

Saitama

How do you get a factor of 2?

$$I_1=\int_0^1 \frac{dt}{\sqrt{1-t^2}(1+e^t)}$$
$$I_3=\int_0^1 \frac{e^t dt}{\sqrt{1-t^2}(1+e^t)}$$

$$I_1+I_3=\int_0^1 \frac{dt}{\sqrt{1-t^2}}$$

How do you get $I_1+I_3=I_2$?

8. Mar 24, 2013

Dick

Look back at the expressions you started with for I1, I2 and I3. I3 integrates $\frac{dt}{\sqrt{1-t^2}(1+e^t)}$ from -1 to 0, I1 integrates from 0 to 1. Put them together and it's the same as integrating -1 to 1. Which is I2.

9. Mar 24, 2013

Saitama

Silly me.

Thanks a lot, Dick!